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- Question
What will be the output of the program?
#include<stdio.h> int main() { float a=0.7; if(a < 0.7) printf("C\n"); else printf("C++\n"); return 0; }
Options- A. C
- B. C++
- C. Compiler error
- D. Non of above
- Correct Answer
- C
Explanationif(a < 0.7) here a is a float variable and 0.7 is a double constant. The float variable a is less than double constant 0.7. Hence the if condition is satisfied and it prints 'C'
Example:#include<stdio.h> int main() { float a=0.7; printf("%.10f %.10f\n",0.7, a); return 0; }Output:
0.7000000000 0.6999999881
Floating Point Issues problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { float a=0.7; if(a < 0.7f) printf("C\n"); else printf("C++\n"); return 0; }
Options- A. C
- B. C++
- C. Compiler error
- D. Non of above Discuss
Correct Answer: C++
Explanation:
if(a < 0.7f) here a is a float variable and 0.7f is a float constant. The float variable a is not less than 0.7f float constant. But both are equal. Hence the if condition is failed and it goes to else it prints 'C++'
Example:#include<stdio.h> int main() { float a=0.7; printf("%.10f %.10f\n",0.7f, a); return 0; }Output:
0.6999999881 0.6999999881- 2. What will be the output of the program?
#include<stdio.h> int main() { float *p; printf("%d\n", sizeof(p)); return 0; }
Options- A. 2 in 16bit compiler, 4 in 32bit compiler
- B. 4 in 16bit compiler, 2 in 32bit compiler
- C. 4 in 16bit compiler, 4 in 32bit compiler
- D. 2 in 16bit compiler, 2 in 32bit compiler Discuss
Correct Answer: 2 in 16bit compiler, 4 in 32bit compiler
Explanation:
sizeof(x) returns the size of x in bytes.
float *p is a pointer to a float.In 16 bit compiler, the pointer size is always 2 bytes.
In 32 bit compiler, the pointer size is always 4 bytes.
- 3. What will be the output of the program?
#include<stdio.h> int main() { float d=2.25; printf("%e,", d); printf("%f,", d); printf("%g,", d); printf("%lf", d); return 0; }
Options- A. 2.2, 2.50, 2.50, 2.5
- B. 2.2e, 2.25f, 2.00, 2.25
- C. 2.250000e+000, 2.250000, 2.25, 2.250000
- D. Error Discuss
Correct Answer: 2.250000e+000, 2.250000, 2.25, 2.250000
Explanation:
printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e+000.printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000.
printf("%g,", d); Here '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25.
printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.
- 4. Which of the following sentences are correct about a switch loop in a C program?
1: switch is useful when we wish to check the value of variable against a particular set of values. 2: switch is useful when we wish to check whether a value falls in different ranges. 3: Compiler implements a jump table for cases used in switch. 4: It is not necessary to use a break in every switch statement.
Options- A. 1,2
- B. 1,3,4
- C. 2,4
- D. 2 Discuss
Correct Answer: 1,3,4
- 5. Which of the following statements are correct about an if-else statements in a C-program?
1: Every if-else statement can be replaced by an equivalent statements using ?: operators 2: Nested if-else statements are allowed. 3: Multiple statements in an if block are allowed. 4: Multiple statements in an else block are allowed.
Options- A. 1 and 2
- B. 2 and 3
- C. 1, 2 and 4
- D. 2, 3, 4 Discuss
Correct Answer: 2, 3, 4
- 6. What will be the output of the program?
#include<stdio.h> #include<math.h> int main() { printf("%f\n", sqrt(36.0)); return 0; }
Options- A. 6.0
- B. 6
- C. 6.000000
- D. Error: Prototype sqrt() not found. Discuss
Correct Answer: 6.000000
Explanation:
printf("%f\n", sqrt(36.0)); It prints the square root of 36 in the float format(i.e 6.000000).Declaration Syntax: double sqrt(double x) calculates and return the positive square root of the given number.
- 7. What will be the output of the program?
#include<stdio.h> #include<math.h> int main() { printf("%d, %d, %d\n", sizeof(3.14f), sizeof(3.14), sizeof(3.14l)); return 0; }
Options- A. 4, 4, 4
- B. 4, 8, 8
- C. 4, 8, 10
- D. 4, 8, 12 Discuss
Correct Answer: 4, 8, 10
Explanation:
sizeof(3.14f) here '3.14f' specifies the float data type. Hence size of float is 4 bytes.sizeof(3.14) here '3.14' specifies the double data type. Hence size of float is 8 bytes.
sizeof(3.14l) here '3.14l' specifies the long double data type. Hence size of float is 10 bytes.
Note: If you run the above program in Linux platform (GCC Compiler) it will give 4, 8, 12 as output. If you run in Windows platform (TurboC Compiler) it will give 4, 8, 10 as output. Because, C is a machine dependent language.
- 8. What will be the output of the program?
#include<stdio.h> #include<math.h> int main() { float n=1.54; printf("%f, %f\n", ceil(n), floor(n)); return 0; }
Options- A. 2.000000, 1.000000
- B. 1.500000, 1.500000
- C. 1.550000, 2.000000
- D. 1.000000, 2.000000 Discuss
Correct Answer: 2.000000, 1.000000
Explanation:
ceil(x) round up the given value. It finds the smallest integer not < x.
floor(x) round down the given value. It finds the smallest integer not > x.
printf("%f, %f\n", ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1.
In the printf("%f, %f\n", ceil(n), floor(n)); statement, the format specifier "%f %f" tells output to be float value. Hence it prints 2.000000 and 1.000000.
- 9. What will be the output of the program?
#include<stdio.h> int main() { float f=43.20; printf("%e, ", f); printf("%f, ", f); printf("%g", f); return 0; }
Options- A. 4.320000e+01, 43.200001, 43.2
- B. 4.3, 43.22, 43.21
- C. 4.3e, 43.20f, 43.00
- D. Error Discuss
Correct Answer: 4.320000e+01, 43.200001, 43.2
Explanation:
printf("%e, ", f); Here '%e' specifies the "Scientific Notation" format. So, it prints the 43.20 as 4.320000e+01.printf("%f, ", f); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 43.20 as 43.200001.
printf("%g, ", f); Here '%g' "Use the shorter of %e or %f". So, it prints the 43.20 as 43.2.
- 10. What will be the output of the program?
#include<stdio.h> int main() { float fval=7.29; printf("%d\n", (int)fval); return 0; }
Options- A. 0
- B. 0.0
- C. 7.0
- D. 7 Discuss
Correct Answer: 7
Explanation:
printf("%d\n", (int)fval); It prints '7'. because, we typecast the (int)fval in to integer. It converts the float value to the nearest integer value.
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- 1. What will be the output of the program?